Physics: Aship leaves a port at noon and travels due west at 20 knots. at 6 pm, a second ship leaves the same port and travels northwest at 15 knots. how fast are the two ships moving apart when the second ship has traveled 90 nautical miles?

Aship leaves a port at noon and travels due west - id3564110, angelaisthebest1700

kaitlyn0123

08.01.2022

v = 12.44 Knots

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

so after time "t" of 2nd ship motion the two ships positions are given as

$r_1 = 20(t + 6)\hat i$

$r_2 = 15(t)(cos45\hat i + sin45\hat j)$

now we can find the distance between two ships as

$x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}$

now we have

x^2 = (120 + 9.4 t)^2 + (10.6 t)^2

x^2 = 200.72 t^2 + 14400 + 2256 t

now we will differentiate it with respect to time

$2x\frac{dx}{dt} = 401.44 t + 2256$

here we know that

$t = \frac{90}{15} = 6 hours$

so we have

x = 187.5

now we have

2(187.5) v = 401.44(6) + 2256

v = 12.44 Knots

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